hdu 5464 Clarke and problem(dp)

Problem Description

Clarke is a patient with multiple personality disorder. One day, Clarke turned into a student and read a book.
Suddenly, a difficult problem appears: 
You are given a sequence of number a1,a2,...,an and a number p. Count the number of the way to choose some of number(choose none of them is also a solution) from the sequence that sum of the numbers is a multiple of p(0 is also count as a multiple of p). Since the answer is very large, you only need to output the answer modulo 109+7

 

Input

The first line contains one integer T(1≤T≤10) - the number of test cases. 
T test cases follow. 
The first line contains two positive integers n,p(1≤n,p≤1000) 
The second line contains n integers a1,a2,...an(|ai|≤109).

 

Output

For each testcase print a integer, the answer.

 

Sample Input

1

2  3

1  2

 

Sample Output

2

  设dp[i][j]表示到第i个元素余数为j的个数。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define modelo 1000000007
int a[1005];
int dp[1005][1005];
int main()
{
    int n,t,q,i,j;
    scanf("%d",&t);
    while (t--)
    {
       scanf("%d%d",&n,&q);
       for (i=1;i<=n;i++)
       {
          scanf("%d",&a[i]);
          a[i]=(a[i]%q+q)%q;
       }
       memset(dp,0,sizeof(dp));
       dp[0][0]=1;
       for (i=1;i<=n;i++)
       {
          for (j=0;j<q;j++)
          dp[i][j]=(dp[i-1][j]+dp[i-1][(j+a[i])%q])%modelo;
       }
       printf("%d
",dp[n][0]);
    }
    return 0;
}